240 9. MEASUREMENTGiven that pyramids are so common in Egypt, it is surprising that the Ahmose
Papyrus does not discuss the volume of a pyramid. However, Problem 14 from the
Moscow Papyrus asks for the volume of the frustum of a square pyramid given that
the side of the lower base is 4, the side of the upper base is 2, and the height is 6.
The author gives the correct recipe: Add the areas of the two bases to the area of
the rectangle whose sides are the sides of the bases, that is, 2 · 2 + 4 · 4 + 2 · 4, then
multiply by one-third of the height, getting the correct answer, 56. This technique
could not have been arrived at through experience. Some geometric principle must
be involved, since the writer knew that the sides of the bases, which are parallel
lines, need to be multiplied. Normally, the lengths of two lines are multiplied only
when they are perpendicular to each other, so that the product represents the area
of a rectangle. Gillings (1972, pp. 190-193) suggests a possible route. Robins
and Shute (1987, pp. 48 49) suggest that the result may have been obtained by
completing the frustum to a full pyramid, and then subtracting the volume of the
smaller pyramid from the larger. In either case, the power of visualization involved
in seeing that the relation is the correct one is remarkable.
Like the surface area problem from the Moscow Papyrus just discussed, this
problem reflects a level of geometric insight that must have required some accu-
mulation of observations built up over time. It is very easy to see that if a right
pyramid with a square base is sliced in half by a plane through its vertex and a
pair of diagonally opposite vertices of the base, the base is bisected along with the
pyramid. Thus, a tetrahedron whose base is half of a square has volume exactly
half that of the pyramid of the same height having the whole square as a base.
It is also easy to visualize how a cube can be cut into two wedges, as in the top
row of Fig. 2. Each of these wedges can then be cut into a pyramid on a face of the
cube plus an extra tetrahedron, as in the bottom row. The tetrahedron P'Q'R'S'
has a base that is half of the square base of the pyramid PQRST, and hence has
half of its volume. It follows that the volume of the tetrahedron is one-sixth that
of the cube, and so the pyramid PQRST is one-third of the volume. A "mixed"
strategy is also possible, involving weighing of the parts. The two tetrahedra would,
in theory, balance one of the square pyramids. This model could be sawn out of
stone or wood. From that special case one might generalize the vital clue that the
volume of a pyramid is one-third the area of the base times the altitude.
Once the principle is established that a pyramid equals a prism on the same
base with one-third the height, it is not difficult to chop a frustum of a pyramid into
the three pieces described in the Moscow Papyrus. Referring to Fig. 3, which shows
a frustum with bottom base a square of side á and upper base a square of side b
with b < a, we can cut off the four corners and replace them by four rectangular
solids with square base of side (a - b)/2 and height h/3. These four fit together to
make a single solid with square base of side a - b and height h/3. One opposite pair
of the four sloping faces that remain after the corners are removed can be cut off,
turned upside down, and laid against the other two sloping faces so as to make a
single slab with a rectangular base that is ï ÷ b and has height h. The top one-third
of this slab can then be cut off and laid aside. It has volume (h/3)ab. The top
half of what remains can then be cut off, and a square prism of side b and height
h/3 cut off from it. If that square prism is laid aside (it has volume (/i/3)6^2 ), the
remaining piece, which is (a - b) χ b ÷ (h/3), will fill out the other corner of the
bottom layer, resulting in a square prism of volume (h/3)a^2. Thus, we obtain the